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Click "Next" button on webpage using selenium in python


By : hannah
Date : September 15 2020, 06:00 PM
I hope this helps you . I have a very simple webpage expecting to click on "NEXT" button after inputting an activation code. But i am not able to find the element by value/name "next" .From the inspect element , This element has name "activationpage1:j_id_id18" so use
code :
driver.find_element_by_name("activationpage1:j_id_id18")
driver.find_element_by_xpath('//input[@name="activationpage1:j_id_id18"]')

driver.find_element_by_xpath('//input[@value="Next >>"]')

driver.find_element_by_xpath('//input[@type="submit"]')

driver.find_element_by_xpath('//input[@style="font-family:Arial,sans-serif;font-size:11px;text-align:center;"]')


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click submit button "OK" on a webpage using python requests module


By : Eleanor Mehlenbacher
Date : March 29 2020, 07:55 AM
hope this fix your issue From the info provided, I assume that you visit a page on http://10.18.25.25/restart/ which has a form, to restart your device on submit.
The action you defined is present at URI /restart/restart on the form. So do a request on same URI:
code :
r = requests.post('http://10.18.25.25/restart/restart', auth=('username', 'password'))

Cannot click "Save" button (selenium webdriver - python - chrome)


By : nepali Sangeet
Date : March 29 2020, 07:55 AM
around this issue If you're using python, the syntax is not right. Python uses snake_case and By uses CONSTANT convention
code :
from selenium import webdriver
from selenium.webdriver.common.by import By

driver.find_element(By.XPATH, "//input[@type='submit' and @value='save']").click()
driver.find_element_by_xpath("//input[@type='submit' and @value='save']").click()
driver.find_element_by_css_selector('input[type="submit"]').click()

Python 3.6 -- Selenium Trying to click on the "Sign in" button on Youtube


By : Cel Pitogo
Date : March 29 2020, 07:55 AM
I think the issue was by ths following , As you were trying to locate the WebElement with text as SIGN IN, the xpath you have used as in browser.find_element_by_xpath('//*[@id="button"]') doesn't identifies the WebElement uniquely. There are 30 such WebElements on the page matching this particular xpath.
Solution :
code :
browser.find_element_by_xpath("//yt-formatted-string[@class='style-scope ytd-button-renderer style-brand']").click()

Python - Selenium: How to click on the element with text as "Year" within the webpage


By : ולרי. ס.
Date : March 29 2020, 07:55 AM
Hope this helps As per the url you have provided to click on the element with text as Year you have to induce WebDriverWait for the element to be clickable as follows:
code :
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.CSS_SELECTOR, "ul.tabs.filter_wrap.clearfix li.ng-scope:nth-child(3)>a"))).click()
WebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.XPATH, "//ul[@class='tabs filter_wrap clearfix']//li[@class='ng-scope']/a[contains(.,'Year')]"))).click()

How to click the "Add friends" button on facebook with selenium python


By : john sullivan
Date : March 29 2020, 07:55 AM
will help you If the xpath provided is correct, try driver.find_element_by_link_text("Add Friend").click() instead of driver.find_elements as driver.find_elements returns a list instead of a single element and that's why it is returning an error in your case.
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